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Hello everyone!
I have found the general formula for:
0011223344...
n - (n - (n - (3 + (-1)^n)/2)/2)
Can any one help me for:
000111222333...
or
0000111122223333....
and so
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0.0011223344
Er, what happens after 99?
0.00112233445566778899 ???
Or does it stop there (so its actually a rational number with a terminating decimal)?
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Wow that's a complicated formula. I think you can simplify it a bit by chopping the front off, if you wanted.
n - (3 + (-1)^n)/2)/2 should work as well.
For the other sequences, I'm not sure. You've made two consecutive terms give the same answer by using (-1)^n, but for more than two that won't work. I'll have a good think about it though, this seems like a fun challenge.
An easy solution would be something like rounddown ((n-1)/3), but presumably you're not allowed to do that kind of thing.
Why did the vector cross the road?
It wanted to be normal.
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Thanks for your replay.
Jane
I think it should work for numbers:
1 2 3 4 5 6 7 8 ........................... n
0 0 1 1 2 2 3 3.......
Correct me if I'm wrong
mathsyperson:
you are right: n - (3 + (-1)^n)/2)/2 works
The formula I found is working for:
001122334455...... only
it is based on finding if the n is odd or even (-1)^n
if I go for 000111222.... or 000011112222....
then I thing the (-1)^n is not valid
Let see for 000111222333...
1 2 3 4 5 6 7 8 9 10 ........n
n
1->0 3/3-1=0 ?
2->0 3/3-1=0 ?
3->0 3/3-1=0
4->1 6/3-1=1 ?
5->1 6/3-1=1 ?
6->1 6/3-1=1
7->2 9/3-1=2 ?
8->2 9/3-1=2 ?
9->2 9/3-1=2
.
.
So it works for n that complitely divides by 3 (no remaining)
So how do I know (by a formula) that a number n complitely divides by 3
or it is: n<3 for 1 and 2
or it is: n<6 for 4 and 5
In fact I did an algorithm for that problem and it works fine:
get the n
see if it complitely devides by 3 then n/3-1
if not n=n+1
then see again if it complitely devides by 3 then n/3-1
and so on
this works but I wanted something beter
like the formula I found for generating 001122334455...
for 1 2 3 4 5 ......n
I'm not that good in maths but I like chalanges,
Maby that information I gave you will do some help.
Thank you again for your replays
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